1
Thermomechanical Reliability
of Microelectronics
Zhen Zhang
School of Engineering and Applied Sciences
Advisor: Zhigang Suo
SRC
MRSEC
2
Topics
Crack penetration or debonding
Interfacial delamination
eEtch Stop
Via
Void
Organosilicate
Liner
500 nm
Dislocation injection
Voiding and electromigraton
Diverse Materials & Sharp Features
Ieong et al, Science 306(2004)p2057.
Intel 90 nm interconnect
Intel Pentium III CPU
Stress concentration
Reuss & Chalamala, MRS bulletin 28(2003)p11.
Failures !
The study of the singular stress filed around
the sharp features draw a lot of attention.
3
Split singularities
Homogeneous material
Two modes have the same
exponents --- 1/2
Bimaterial
Two modes have different
exponents --- λ1 and λ2
K
K
σ ij (r ,θ ) = 1/I2 Σ ijI (θ ) + 1/II2 Σ ijII (θ )
r
r
k1 1
k2 2
σ ij (r ,θ ) = λ1 Σ ij (θ ) + λ2 Σ ij (θ )
r
r
4
5
Split due to geometry
Singularity exponents λ
1
0.75
degeneracy
λ1
0.5
Split
0.25
λ2
0
0
30
60
90
120
Wedge angle ω (deg)
150
180
Liu, Suo and Ma, Acta mat. (1999).
6
Split due to elastic mismatch
1
Singularity exponents λ
0.75
0.5
λ1
0.25
λ2
0
−2
10
−1
10
0
10
Modulus ratio E1/E2
1
10
2
10
Always split, unless Mat.#1 is rigid.
Split singularities: oblique crack
1
β=0
0.9
α=
singularity exponents λ
0.8
E1 − E2
E1 + E2
λ1
α = −0.9
1 ⎡ µ1 (1 − 2ν 2 ) − µ 2 (1 −λ22ν 1 ) ⎤
β= ⎢
2 ⎣ µ1 (1 − ν 2 ) + µ 2 (1 − ν 1 ) ⎥⎦
0.7
0.6
λ1
0.5
α = 0.9
0.4
0.3
λ2
0.2
0.1
0
0
10
20
30
Mat. #1
Mat. #2
ω
40
50
60
impinging angle ω
70
80
90
Mat. #1
Mat. #1
Mat. #2
Mat. #2
7
8
Split singularities:
Rule rather than exception:
k1 1
k2 2
σ ij (r , θ ) = λ1 Σ ij (θ ) + λ2 Σ ij (θ )
r
r
Questions:
• Should we neglect the weaker singularity?
• How do we study the failure phenomena with
consideration of weaker singularity?
9
Applications and Implications: #1
Crack Penetration or Debonding
Cracking Path Selection
k1, k2
A crack impinges upon an interface.
(He and Hutchinson, IJSS, 1989).
Penetration
vs.
ω
Mat. #1
Mat. #2
Debonding
The effect of weaker singularity has not been well studied.
10
Q: Should we neglect the weaker singularity?
Homogeneous material:
k-annulus
r
KI I
K II II
σ ij (r , θ ) = 1/ 2 Σ ij (θ ) + 1/ 2 Σ ij (θ )
r
r
K II
Mode mixity =
KI
Bimaterial:
k1 1
k2 2
σ ij (r , θ ) = λ1 Σ ij (θ ) + λ2 Σ ij (θ )
r
r
Stress ratio
of two modes
r does NOT go to zero.
=
k 2 r − λ2
k1r −λ1
Selection of arbitrary length
11
12
Definition of mode mixity
k1 1
k2 2
σ ij (r , θ ) = λ1 Σ ij (θ ) + λ2 Σ ij (θ )
r
r
k-annulus
r
Select some length we care:
k 2 Λ− λ2
Mode mixity: η =
k1Λ−λ1
It measures the relative
contribution of the two modes to
the stress field at length scale Λ
Answer: it depends on the magnitude of the mode mixity.
13
Two length scales
k1 = κ1TL ~ [stress ][length]
λ1
λ1
k 2 = κ 2TL ~ [stress][length]
λ2
λ2
Mat. #1
Λ = 1 nm
Mat. #2
(
(
Im Kl iε
Re Kl iε
)
)
(Rice, 1988)
κ2 ⎛ Λ ⎞
η= ⎜ ⎟
κ1 ⎝ L ⎠
λ1 − λ2
Mat. #1
ω
Mat. #2
L = 100 nm
0.2
κ2 ⎛ Λ ⎞
η= ⎜ ⎟
κ1 ⎝ L ⎠
λ1 − λ2
0.4
easily on the order of unity.
η
is used to assess the effect of weaker singularity.
14
Crack penetration
k1, k2
Mat. #1
ω
Mat. #2
90
Mode angle of the penetrating crack, ψ p (deg)
(a)
α = −0.5, β = 0, ω = 45
κ ⎛a⎞
η= 2⎜ ⎟
κ1 ⎝ L ⎠
60
30
k2
K Ip
k1
= b11 ⋅ λ1 + b12 ⋅ λ2
a
a
a
0
λ1 − λ2
k
K IIp
k
= b21 ⋅ λ11 + b22 ⋅ λ22
a
a
a
η = 10
η=1
0
η=0
−30
Selection of penetrating angle:
η = −1
−60
K IIp = 0
η = −10
−90
0
30
60
90
120
Penetrating angle, φ (deg)
150
180
Penetrating angle vs. mode mixity
Penetrating angle φ* (deg) corresponding to KpII = 0
Compliant
× Hayashi and Nemat-Nasser (1981)
180
o He and Hutchinson (1988,1989)
150
120
15
Stiff
α = −0.5
+
α=0
90
α = 0.5
60
Homogenous
*
30
0
β = 0, ω = 45
0
−3
−2
Stiff
+
−1
0
Local mode mixity η
1
2
3
Weaker singularity plays role via mode mixity.
Compliant
16
Debonding
k1, k2
Mat. #1
ω
Mat. #2
90
κ2 ⎛ a ⎞
η= ⎜ ⎟
κ1 ⎝ L ⎠
60
λ1 − λ2
Mode angle of debond crack, ψ
d
(deg)
β = 0, ω = 450
K Id
k
k
= c11 ⋅ λ11 + c12 ⋅ λ22
a
a
a
30
0
K IId
k1
k2
= c21 ⋅ λ1 + c22 ⋅ λ2
a
a
a
−30
α = 0.5
α=0
α = −0.5
−60
−90
−3
−2
−1
0
Local mode mixity, η
1
2
3
17
Penetration vs. Debonding
(
(
)
)
(
(
)
)
2
2
c112 + c21
+ 2(c11c12 + c21c22 )η + c122 + c22
η2
1
=
⋅ 2
p
1 − α b11 + b212 + 2(b11b12 + b21b22 )η + b122 + b222 η 2
G
Gd
Γ1
1
0
Ratio of energy release rates G d / G p
β = 0, ω = 45
Γi
Γ1
0.75
Γi
Gd
<
If
, Penetrate.
p
G
Γ1
penetration
debonding
0.5
α = −0.5
Γi (ψ d )
0.25
α=0
α = 0.5
0
−2
−1
0
Local mode mixityλη1 − λ2
η=
κ2 ⎛ a ⎞
⎜ ⎟
κ1 ⎝ L ⎠
1
2
Γi
Gd
If G p > Γ , Debond.
1
Weaker singularity can readily change the outcome of the penetration-debond competition
18
Applications and Implications: #2
Interfacial Delamination
due to
Chip-Package Interaction
19
Multi-level structure of flip-chip
Underfill
(~50-100 µm thick)
Stress
concentration
Delamination
Silicon Die
( ~0.7mm thick, ~10-15mm wide)
Solder joints
(~50 µm diameter)
Package Substrate
(~1mm thick, ~3-5cm wide)
Interconnects
Etch stop
Passivation
`
Level #2
Delamination
Via
Metal level #1
Dielectrics
~ 50µm
~100nm
Liner
underfill/solder
joint cracking
20
Observations:
• Failure mode: interfacial delamination
• Driving force: chip-packaging interaction
Challenges:
•
•
•
•
Huge variation in length scales: ~nm to ~cm
3D multilevel structures
Diverse materials interaction
Global-local FEM is adopted in industry.
21
Simplified model
k-field
σ ij (r ,θ ) =
k1 1
k2 2
(
)
Σ
+
Σ ij (θ )
θ
ij
λ1
λ2
r
r
Delamination
Etch stop
Level #2
Via
Metal level #1
Dielectrics
~100nm
Liner
• k-field scales with h.
• Small feature size is within
k-field.
• Same macroscopic driving
force motivates different
flaws to grow.
• The same flaw size and
orientation, the same ERR.
• Maybe different toughness.
• Local driving force is much
smaller than global driving
force.
22
k-field and K-field
Chip-package interaction:
k1 1
k2 2
σ ij (r , θ ) = λ1 Σ ij (θ ) + λ2 Σ ij (θ )
r
r
k1, k2
Interfacial delamination:
K
(
)
(
)
k
k
Re Ka iε
= c11 ⋅ λ11 + c12 ⋅ λ22
a
a
a
k1
k2
Im Ka iε
= c21 ⋅ λ1 + c22 ⋅ λ2
a
a
a
23
Length dependence of mode mixity
1− 2 λ1
1 − β 2 κ12σ 2 h ⎛ a ⎞
G=
⎜ ⎟
1 − α ESi ⎝ h ⎠
[(c
11 + c12η ) + (c21 + c22η )
2
2
]
κ2 ⎛ a ⎞
η= ⎜ ⎟
κ1 ⎝ h ⎠
λ1 − λ2
Thermal excursion ∆T=1400C
2
Energy release rate, G (J/m2)
10
κ ⎛a⎞
η= 2⎜ ⎟
κ1 ⎝ h ⎠
λ1 − λ2
Wang et.al. (2003)
1
10
x
chip/FR4
chip/LTCC
0
10 −4
10
−3
−2
−1
10
10
10
Normalized crack length, a/h
0
10
• Similar value as multiscale calculation.
• k-field applies if
a/h<1/4.
• Break-down of power
law of G~a relation.
• G hardly goes to zero.
24
Reduction in singularity
Silicon die
Filler/coating
Polymer substrate
165 GPa
23 GPa
λ1 = 0.497
λ2 = 0.346
Die
Polymer substrate
10 GPa
165 GPa
23 GPa
λ1 = 0.3
λ2 = 0.247
Singularities are reduced a lot by using a filler or coating.
Why?
Elastic mismatch is reduced and geometric discontinuity is closed.
25
Suppression of debonding
Silicon die
Die
Polymer substrate
Free surfaces are easy to deform,
interfacial flaw is easy to open.
Polymer substrate
The opening of the debonded
interface is suppressed.
Example in flexible electronics
Coating
Island
Normalized energy release rate, G / ( E*f ε20 h )
Substrate
• Enhance the survival rate
• Increase the island size
0.3
0.2
0.1
Ef / Es = 40
L / h = 100
+
a / h = 0.1
0
−3
10
−2
−1
0
1
2
10
10
10
10
10
Moduli ratio of coating to substrate, Ec / Es
3
10
26
27
Applications and Implications: #3
Dislocation Injection
From Sharp Features
In Strained Silicon Structures
28
Strained silicon structure
• Stresses are deliberately
introduced to increase
carrier mobility.
• Edges and corners intensify
stresses, inject dislocations
and fail the device.
• Objective: critical
conditions that avert
dislocations, on the basis of
singular stress fields near
the sharp features.
29
Simplification
k1 1
k2 2
σ ij (r ,θ ) = λ1 Σ ij (θ ) + λ2 Σ ij (θ )
r
r
λ1 = 0.4514
κ ⎛b⎞
η= 2⎜ ⎟
κ1 ⎝ h ⎠
λ1 − λ2
λ2 = 0.0752
b = 0.383 nm
h = 100 nm
η ~ 0.02
k
σ ij (r ,θ ) = λ Σ ij (θ )
r
30
Stress intensity factor k
k = σh f (L / h )
λ
k-annulus
• The driving force of
dislocation injection is
stress field in k-annulus,
which is characterized by k.
Normalized stress intensity factor, k / ( σ hλ)
Critical stress intensity factor kc
31
0.5
0.4
Critical condition:
k = σh f (L / h )
λ
k = kc
0.3
0.2
Driving force
0.1
0
0
5
10
15
20
25
Aspect ratio of SiN stripe, L / h
Resistance
30
kc is specific to the materials and wedge angle,
like fracture toughness.
32
An estimate of kc
σ ij (r , θ ) =
k
Σ ij (θ )
λ
r
τ nb = σ ij ni b j
τ th =
µb
≅ 0.2µ
2πd
Criterion:
τ nb r =b = τ th
k c = 0 .5 µ b
1
2
(111)[10 1 ]
1
2
(111)[01 1 ]
Isomae (1981)
λ
33
Critical stress in SiN stripe
0.5µ ⎛ b ⎞
σc =
⎜ ⎟
f (L / h ) ⎝ h ⎠
λ
The wider and thicker,
The easier to inject dislocation.
0.12
Normalized critical stress , σc / µ
0.11
0.1
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0
5
10
15
20
Aspect ratio of SiN pad, L / h
25
30
Kammler et al (2005)
34
Measurement of kc
Calculate k:
Measure kc :
k = σh λ f (L / h )
35
Summary
• Split singularities:
k1 1
k2 2
σ ij (r ,θ ) = λ1 Σ ij (θ ) + λ2 Σ ij (θ )
r
r
• Mode mixity depends on two lengths:
⎛ κ 2 ⎞⎛ Λ ⎞
η = ⎜⎜ ⎟⎟⎜ ⎟
⎝ κ1 ⎠⎝ L ⎠
λ1 − λ2
36
Acknowledgements
• Advisor:
• Committee:
Zhigang Suo
John Hutchinson
Joost Vlassak
Frans Spaepen
• Collaborators: Juil Yoon, Nanshu Lu, Martijn
Feron, Jun He, Jim Liang, Jean Prevost, etc.
• Group members: Wei Hong, Xuanhe Zhao,
Jinxiong Zhou, Dunming Liao, etc.
37
Backup slides for split
singularities
38
Split singularties
0.5
0.15
0.1
0.4
0.4
0.25
0.51
0.05
0.8
0.1
0.2
0.49
0.5
0.6
0.7
0.3
0.4
0.51
0.5
0.5 0.49
0.6
0.7
β
0
0.51
0.51 0.5 0.49
0.49
0
α
0.25
0.05
−0.25
0.6
−0.5
−1
−0.75
−0.5
−0.25
0.5
0.75
1
39
0.5
0.1
0.2
0.3
0.25
0.5
0
0.7
0.8
0.6
0.9
β
0.4
−0.25
−0.5
−1
−0.75
−0.5
−0.25
0
α
0.25
0.5
0.75
1
0.15
40
0.05
0.45
0.49
0.25
0.1
0.5
0.5
0.49
0.45
0.2
0
−0.25
0.3
0.05
0.1
0.4
0.01
β
0.45
0.4
0.3
0.49
0.2
−0.5
−1
0.1
−0.75
−0.5
−0.25
0
α
0.25
0.5
0.75
1
41
0.5
Singularity exponent, λ
Re (λ)
0.4
0.3
λ1
0.2
0.1
0
−2
10
λ2
Im (λ)
−1
0
1
10
10
10
Moduli ratio of coating to substrate, Ec / Es
2
10
42
k
K Ip
k
= b11 ⋅ λ11 + b12 ⋅ λ22
a
a
a
k2
K IIp
k1
= b21 ⋅ λ1 + b22 ⋅ λ2
a
a
a
43
Backup slide for strained silicon
44
Measurement of kc
A series of experiments:
Dimensional requirement:
kc = c ⋅ µ b
k = σh λ f (L / h )
λ
Find a critical condition to obtain coefficient c by:
λ
σ ⎛h⎞
⎛L⎞
c = ⋅⎜ ⎟ ⋅ f ⎜ ⎟
µ ⎝b⎠
⎝h⎠
45
Other concerns from industry
• Periodic patterned, so how is the spacing
effect?
• 3D corner is more singular than 2D wedge.
Does the current method apply?
• What if I don’t use SiN? I chose other
materials as stressors.
46
Spacing effect
k
σ ij (r ,θ ) = λ Σ ij (θ )
r
with
⎛L S⎞
k = σh ⋅ f ⎜ , ⎟
⎝h h⎠
Normalized stress intensity factor, k / ( σ hλ)
0.6
λ
L/h=20
0.5
0.4
0.3
0.2
λ = 0.4514
In practice, S/h doesn’t go
to zero, so the spacing
effect is quite small.
L/h=2
0
5
10
15
Ratio of the spacing to the thickness of pads, S / h
20
47
3D Corners
⎛L S R⎞
k = σh ⋅ f ⎜ , , ⎟
⎝h h h⎠
λ
• The method still applies.
What changes is the
coefficients.
• 3D corner singularities.
Measurement of kc and
calculation of f(L/h,
S/h, R/h).
• 3D calculation is
expensive for academia,
but very cheap for
industry. Do it.
48
Effect of weaker singularity if not SiN
κ2 ⎛ b ⎞
Local mode mixity: η = ⎜ ⎟
κ1 ⎝ h ⎠
λ1 − λ2
1
1
L/h = 2
0.8
0.6
Mode mixity, η
Singularity exponents λ
0.75
0.5
λ1
L/h = 20
0.4
0.2
0.25
0
λ2
0
−2
10
−1
10
0
10
Modulus ratio E1/E2
1
10
2
10
−0.2
−2
10
−1
10
Evaluate it case by case.
0
10
Modulus ratio, E1/E2
1
10
2
10
49
Slip systems
50
Future focus
Ge et al, IEDM 2003.
• Stress driven dislocation motion under channel.